2
$\begingroup$

Prove that $\phi_a(z)=\frac{a-z}{1-\bar az}$ , $0<|a|<1$ has exactly two fixed points ; one inside the unit disc and the other outside the unit disc.

Putting $\phi_a(z)=z$ I find that there are exactly two fixed points which are $$\frac{1\pm \sqrt{1-|a|^2}}{\bar a}.$$

I am unable to find out the second part.

Let $z_1=\frac{1- \sqrt{1-|a|^2}}{\bar a}$. From $0<|a|<1$ , we get $z_1<\frac{1}{\bar a}=\frac{a}{|a|^2}$. Then , $|z_1|<\frac{1}{|a|}$. But how I show that $|z_1|<1$ so that I can say that $z_1$ lies inside the circle ?

$\endgroup$
2
$\begingroup$

You are solving the quadratic $$\def\abar{\overline a}\abar z^2-2z+a=0\ .$$ By the usual product-of-roots formula we have $$z_1z_2=a\,/\,\abar$$ and so $$|z_1||z_2|=|a\,/\,\abar|=1\ .$$ Therefore one of $z_1$ and $z_2$ has modulus less than $1$, the other greater.

(It could be that both have modulus equal to $1$, but it is not very hard to see that this is impossible.)

$\endgroup$
1
  • $\begingroup$ You beat me to the punch line! +1 $\endgroup$
    – Mark Viola
    Jul 20 '15 at 3:26
0
$\begingroup$

$z_1\overline{a}=1-\sqrt{1-x^2}, x = |a|\to (|z_1||\overline{a}|)^2 =1-2\sqrt{1-x^2}+1-x^2\to bx^2= 2-x^2-2\sqrt{1-x^2}$. Thus we can prove: $2-x^2-2\sqrt{1-x^2} < x^2 \iff 1-x^2 < \sqrt{1-x^2}$ which is true since $0 < 1-x^2 < 1$. Thus $b = |z_1| < 1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.